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HMMT February

Algebra Difficulty 5.2 AIME, harder Find the answer

Let a0=−2,b0=1a_{0}=-2, b_{0}=1, and for n≥0n \geq 0, let an+1=an+bn+an2+bn2bn+1=an+bn−an2+bn2\begin{aligned} & a_{n+1}=a_{n}+b_{n}+\sqrt{a_{n}^{2}+b_{n}^{2}} \\ & b_{n+1}=a_{n}+b_{n}-\sqrt{a_{n}^{2}+b_{n}^{2}} \end{aligned} Find a2012a_{2012}.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

We have an+1+bn+1=2(an+bn)an+1bn+1=(an+bn)2−an2−bn2=2anbn\begin{gathered} a_{n+1}+b_{n+1}=2\left(a_{n}+b_{n}\right) \\ a_{n+1} b_{n+1}=\left(a_{n}+b_{n}\right)^{2}-a_{n}^{2}-b_{n}^{2}=2 a_{n} b_{n} \end{gathered} Thus, an+bn=−2nanbn=−2n+1\begin{aligned} a_{n}+b_{n} & =-2^{n} \\ a_{n} b_{n} & =-2^{n+1} \end{aligned} Using Viete's formula, a2012a_{2012} and b2012b_{2012} are the roots of the following quadratic, and, since the square root is positive, a2012a_{2012} is the bigger root: x2+22012x−22013x^{2}+2^{2012} x-2^{2013} Thus, a2012=2100622010+2−22011a_{2012}=2^{1006} \sqrt{2^{2010}+2}-2^{2011}

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